Prove that if the real-valued function f is strictly increasing, then f is oneto-one.

See proof below

Step-by-step explanation:

Let x,y be arbitrary real numbers. We want to prove that if x≠y then f(x)≠f(y) (this is the definition of 1-1).

If x≠y, we can assume, without loss of generality that x<y using the trichotomy law of real numbers (without loss of generality means that the argument in this proof is the same if we assume y<x).

Because f is strictly increasing, x<y implies that f(x)<f(y). Therefore f(x)≠f(y) because of the trichotomy law, and hence f is one-to-one.

what are ur choices?

option: d is the correct answer.

d) none of the above.

step-by-step explanation:

clearly from the figure attached to the question we could see that there is no shape given. hence without any shape given in the figure and after rotation we can't obtained a square pyramid.

since, to obtain a image we need a proper figure of the pre-image so that it can be transformed.

hence, the correct answer is:

d) none of the above.

step-by-step explanation: